Using the Prime Number Theorem show that:

$$x^{\pi(x)} < 3^x$$ for sufficiently large $x$.

I started off by taking the $\log$ of the inequality such that:

$$\log(x^{\pi(x)}) < \log(3^x)$$ so $$\pi(x) \log(x) < x \log(3)$$

by PNT we know that $$\pi(x) \sim \dfrac{x}{\log(x)}$$

so can we then just sub this in and obtain:

$$\dfrac{x}{\log(x)} \log(x)<x\log(3)$$ $$x<x\log(3)$$ $$1<\log(3)$$

Can I do this? This is for a Distribution of Primes course, I'm trying to get my head around using the PNT! Thanks for any help!

  • $\begingroup$ Please. Write "a\log b", not "a log b". The former gives you $a\log b$, the latter gives you $a log b$. I also changed \text{~} to \sim, thus $\pi(x)\text{~}\dfrac{x}{\log(x)}$ became $\pi(x)\sim\dfrac{x}{\log(x)}$. ${}\qquad{}$ $\endgroup$ – Michael Hardy May 21 '14 at 16:55
  • $\begingroup$ Your reasoning leads to the conclusion that $x^{\pi(x)} < 3^x$ if $x$ is sufficiently large. But how large is sufficiently large must be examined. $\endgroup$ – Michael Hardy May 21 '14 at 16:57

You need to be a bit careful when handling the $\sim$ relation. You can't just substitute it in the inequality, since it is only an asymptotic relation. This is one way to do it:

The $\sim$ operator is usually defined as

$$f(x)\sim g(x) \iff \lim_{x \to \infty} \frac{f(x)}{g(x)}=1$$

So, $$\pi(x) \sim \frac x{\log x} \iff \lim_{x \to \infty}\frac{\pi(x) \log x}{x}= 1 $$

Dividing the inequality $\pi(x)\log x < x \log 3$ by $x$ gives you $$\frac{\pi(x)\log x}x < \log 3$$

Finally, by taking the limit to infinity of the left hand side you can conclude that the value will approach $1$ for a large enough $x$, and you are done, since $3 > e \implies \log 3 >1$.

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  • $\begingroup$ Ah yes, I forgot I need to use the limit bit! Thank you very much! $\endgroup$ – sarahusher May 21 '14 at 16:54

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